Question

Question 13 Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = (m ± tanθ) / (1 ∓ mtanθ)

Class X1 - Maths -Straight Lines Page 234

Answer 1

concept :- angle between two lines of slope m1 and m2 is tan∅ =|m2 - m1|/|1 + m1.m2|

slope of given line is m .
Let slope of another line is m2.
now,
angle between them tan∅ = |m - m2|/|1 + m.m2|
± tan∅ = (m - m2/(1 + m.m2)

take positive sign ,
tan∅ = (m - m2)/(1 + m.m2)
(1 + m.m2)tan∅ = (m - m2)
tan∅ + m.m2.tan∅ = m - m2
m.m2.tan∅ + m2 = m - tan∅
m2(m.tan∅ + 1) = ( m - tan∅)
m2 = ( m - tan∅)/(m.tan∅ + 1) _____(1)

taking negative sign,
-tan∅ = (m - m2)/(1 + m.m2)
-tan∅ - m.m2.tan∅ = m - m2
m2 - m.m2.tan∅ = m + tan∅
m2( 1 - m .tan∅) = ( m + tan∅)
m2 = ( m + tan∅)/(1 - m.tan∅)_______(2)

from equations (1) and (2)
m2 = (m ± tan∅) / (1 ∓ mtan∅)________(3)

but equation of line passing through origin(0,0)
so, ( y - 0) = m2(x - 0)
y/x = m2
from equation (3)
y/x = (m ± tan∅) / (1 ∓ mtan∅)

hence proved

Answer 2

Answer:

let the equation of the line passing through the origin be y=m1x.

if this line make an angle of theata with line y=mx+c,then angle theata is given by

tan theata = (m1-m/1+m1m)

= (y/x-m/1+y/xm)

=±(y/x-m/1+y/xm)

=y/x-m/1+y/xm

case 1; tan theta = y/x-m/1+y/xm

tan theata +y/x-m tan theata =y/x-m

m+tan theata = y/x(1-m tan theata)

y/x=m+tan theata/1-m tan theata

case 2: tan theata =-{y/x-m/1+y/xm}

=tan theata+y/xm tan theata = -y/x+m

= y/x(1+m tan theata) = m-tan theata

=y/x=m-tan theata/1+m tan theata

therefore, y/x=m±tan theata/1±m tan theata